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Riyadh Philanthropic Society For Science Prince Sultan College For Woman

Riyadh Philanthropic Society For Science Prince Sultan College For Woman Dept. of Computer & Information Sciences CS 340 Introduction to Database Systems (Chapter 10 Tutorial). Exercise 1 Consider a relation R(A, B, C, D, E) with the following dependencies: AB C CD E DE B

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Riyadh Philanthropic Society For Science Prince Sultan College For Woman

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  1. Riyadh Philanthropic Society For Science Prince Sultan College For Woman Dept. of Computer & Information Sciences CS 340 Introduction to Database Systems (Chapter 10 Tutorial)

  2. Exercise 1 • Consider a relation R(A, B, C, D, E) with the following dependencies: • AB C CD E DE B • Is AB a candidate key of this relation? No, AB+ = {A, B, C}. If not, is ABD? Yes, ABD+ = {A, B, C, D, E}. Chapter 10 Tutorial 1

  3. Exercise 2 • Consider the following relation: • CAR_SALE(Car#, DateSold, Salesman#, Commission%, DiscountAmount) • Assume that a car may be sold by multiple salesmen, and hence {Car#, Salesman#} is the primary key. Additional dependencies are: • Car#  DateSold • Car#  DiscountAmount • DateSold  DiscountAmount • Salesman#  Commission% • Based on the given primary key, is the relation in 1NF, 2NF, 3NF? • Why or why not? • How would you successively normalize it completely? Chapter 10 Tutorial 2

  4. Exercise 2 • The relation is in 1NF because all attribute values are single atomic • values. • The relation is not in 2NF because: • Car#  DateSold • Car#  DiscountAmount • Salesman#  Commission% • Thus, these attributes are not fully functionally dependent on the • primary key. • 2NF decomposition: • CAR_SALE1(Car#, DateSold, DiscountAmount) • CAR_SALE2(Car#, Salesman#) • CAR_SALE3(Salesman#, Commission%) Chapter 10 Tutorial 3

  5. Exercise 2 • The relations are not in 3NF because: • Car#  DateSold  DiscountAmount • Thus, DateSold is neither a key itself nor a subset of a key and • DiscountAmount is not a prime attribute. • 3NF decomposition: • CAR_SALES1A(Car#, DateSold) • CAR_SALES1B(DateSold, DiscountAmount) • CAR_SALE2(Car#, Salesman#) • CAR_SALE3(Salesman#, Commission%) Chapter 10 Tutorial 4

  6. Exercise 3 • Consider the following relation for published books: • BOOK(BookTitle, AuthorName, BookType, ListPrice, AuthorAffiliation, Publisher) • Suppose the following dependencies exist: • BookTitle  BookType, Publisher • BookType  ListPrice • AuthorName  AuthorAffiliation • What normal form is the relation in? explain your answer. • Apply normalization until you cannot decompose the relations further. State the reasons behind each decomposition. Chapter 10 Tutorial 5

  7. Exercise 3 • The relation is in 1NF and not in 2NF as no attributes are fully • functionally dependent on the key (BookTitle and AuthorName). It • is also not in 3NF. • The relation is not in 2NF because: • BookTitle  Publisher, BookType • BookType  ListPrice • AuthorName  AuthorAffiliation • Thus, these attributes are not fully functionally dependent on the • primary key. The 2NF decomposition will eliminate the partial • dependencies. • 2NF decomposition: • Book1(BookTitle, AuthorName) • Book2(BookTitle, BookType, ListPrice, Publisher) • Book3(AuthorName, AuthorAffiliation) Chapter 10 Tutorial 6

  8. Exercise 3 • The relations are not in 3NF because: • BookTitle BookType ListPrice • Thus, BookType is neither a key itself nor a subset of a key and ListPrice is not a prime attribute. The 3NF decomposition will • eliminate the transitive dependency of Listprice. • 3NF decomposition: • Book1(BookTitle, AuthorName) • Book2A(BookTitle, BookType, Publisher) • Book2B(BookType, ListPrice) • Book3(AuthorName, AuthorAffiliation) • The relations are also in BCNF. Chapter 10 Tutorial 7

  9. Exercise 4 • Consider a relation R(A, B, C, D, E) with the following dependencies: • AB  C • B  D • C  B • A  E • a. Specify the key of R. • b. Decompose the relation into a set of 3NF relations. • c. Decompose the relation into a set of BCNF relations. • d. Why might one want to stop at 3NF and not decompose further to BCNF. Chapter 10 Tutorial 8

  10. Exercise 4 • a. Specify the key of R. • A and B • b. Decompose the relation into a set of 3NF relations. • R1(A, B, C) • R2(B, D) • R3(A, E) • c. Decompose the relation into a set of BCNF relations. • R1(A, C) • R2(C, B) • R2(B, D) • R3(A, E) Chapter 10 Tutorial 9

  11. Exercise 4 • d. Why might one want to stop at 3NF and not decompose further to • BCNF. • The reason that 3NF is sometimes preferred is that it allows one to • preserve each FD in a minimal cover of FDs within one relation, • which allows these constraints to be checked more easily. Chapter 10 Tutorial 10

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